338. Counting Bits【M】【leetcode题解】

Given a non negative integer number num. For every numbers i in the range 0 ≤ i ≤ num calculate the number of 1's in their binary representation and return them as an array.

Example:
For num = 5 you should return [0,1,1,2,1,2].

Follow up:

• It is very easy to come up with a solution with run time O(n*sizeof(integer)). But can you do it in linear time O(n) /possibly in a single pass?
• Space complexity should be O(n).
• Can you do it like a boss? Do it without using any builtin function like __builtin_popcount in c++ or in any other language.

Show Hint 

1. Credits:
   Special thanks to @ syedee for adding this problem and creating all test cases.

   
   

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   它是有规律地重复的

   
   

   
   

   
   

   
   

   class Solution(object):
       def countBits(self, num):
           if num == 0:
               return [0]
           res = [0]
           bit = 0
           temp = num
           while temp > 0:
               temp /= 2
               bit += 1
           #print bit
           for i in xrange(bit-1):
               t_res = []
               for j in res:
                   t_res.append(j+1)
               res = res + t_res
               #print res
           for i in res[:(num - pow(2,bit-1))+1]:
               res.append(i+1)
   
           return res